Consider the polar curve $r=\sin(\theta)\cos(\theta)$. What is the equation of the tangent line to the curve $r$ at $\theta=\dfrac{\pi}{3}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-\dfrac{3}{4}=\dfrac{\sqrt{3}}{4}\left(x-\dfrac{\sqrt{3}}{4}\right)$ (Choice B) B $y-\dfrac{3}{8}=\dfrac{\sqrt{3}}{4}\left(x-\dfrac{\sqrt{3}}{8}\right)$ (Choice C) C $y-\dfrac{3}{4}=\dfrac{\sqrt{3}}{5}\left(x-\dfrac{\sqrt{3}}{4}\right)$ (Choice D) D $y-\dfrac{3}{8}=\dfrac{\sqrt{3}}{5}\left(x-\dfrac{\sqrt{3}}{8}\right)$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{\pi}{3}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\sin(\theta)\cos(\theta)}\cos(\theta) \\\\ &=\sin(\theta)\cos^2(\theta) \\\\ y&={\sin(\theta)\cos(\theta)} \sin(\theta) \\\\ &=\sin^2(\theta)\cos(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{2\sin(\theta)\cos^2(\theta)-\sin^3(\theta)}{\cos^3(\theta)-2\sin^2(\theta)\cos(\theta)} \end{aligned}$ Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{3}}$ gives us the slope of our tangent line. $\begin{aligned} {\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{\pi}{3}}} &=\dfrac{2\sin\left({\dfrac{\pi}{3}}\right)\cos^2\left({\dfrac{\pi}{3}}\right)-\sin^3\left({\dfrac{\pi}{3}}\right)}{\cos^3\left({\dfrac{\pi}{3}}\right)-2\sin^2\left({\dfrac{\pi}{3}}\right)\cos\left({\dfrac{\pi}{3}}\right)} \\\\ &=\dfrac{2\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{3}}{2}\right)^3}{\left(\dfrac{1}{2}\right)^3-2\left(\dfrac{\sqrt{3}}{2}\right)^2\left(\dfrac{1}{2}\right)} \\\\ &=\dfrac{2\sqrt{3}-3\sqrt{3}}{1-6} \\\\ &={\dfrac{\sqrt{3}}{5}} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{\pi}{3}$. $\begin{aligned} {x\left({\dfrac{\pi}{3}}\right)}&=\sin\left({\dfrac{\pi}{3}}\right)\cos^2\left({\dfrac{\pi}{3}}\right) \\\\ &=\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)^2 \\\\ &={\dfrac{\sqrt{3}}{8}} \\\\ \\\\ y\left({\dfrac{\pi}{3}}\right)}&=\sin^2\left({\dfrac{\pi}{3}}\right)\cos\left({\dfrac{\pi}{3}}\right) \\\\ &=\left(\dfrac{\sqrt{3}}{2}\right)^2\left(\dfrac{{1}}{2}\right) \\\\ &=\dfrac{{3}}{8}} \end{aligned}$ Therefore the equation of our tangent line is: $\begin{aligned} y-\dfrac{3}{8}}&={\dfrac{\sqrt{3}}{5}}\left(x-{\dfrac{\sqrt{3}}{8}}\right) \end{aligned}$ The graph of the tangent is shown. ${0.4}$ ${0.8}$ ${1.2}$ ${0}$ ${\frac{1}{6}\pi}$ ${\frac{1}{3}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{2}{3}\pi}$ ${\frac{5}{6}\pi}$ ${\pi}$ ${\frac{7}{6}\pi}$ ${\frac{4}{3}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{5}{3}\pi}$ ${\frac{11}{6}\pi}$